Question

tan 9 theta - tan 5 theta - tan 4 theta = tan 9 theta tan5 theta tan 4 theta​

Answer 1

$\large\underline{\sf{Solution-}}$

We have to prove that

$\rm \: tan9\theta - tan5\theta - tan4\theta = tan9\theta tan5\theta tan4\theta$

To Prove this, we know that

$\rm \: 9\theta = 5\theta + 4\theta$

So,

$\rm \: tan9\theta = tan(5\theta + 4\theta)$

We know

$\boxed{\sf{ \: \: tan(x + y) \: = \: \frac{tanx \: + \: tany}{1 \: - \: tanx \: tany \: } \: \: }}$

So, using this identity, we get

$\rm \: tan9\theta = \dfrac{tan5\theta + tan4\theta }{1 - tan5\theta \: tan4\theta }$

$\rm \: tan9\theta - tan9\theta tan5\theta tan4\theta = tan5\theta + tan4\theta$

So, on rearranging the terms, we get

$\rm \: tan9\theta - tan5\theta - tan4\theta = tan9\theta tan5\theta tan4\theta$

Hence,

$\boxed{\sf{ \:\rm \: tan9\theta - tan5\theta - tan4\theta = tan9\theta \: tan5\theta \: tan4\theta \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

$\boxed{\sf{ \:sin(x + y) = sinx \: cosy \: + \: siny \: cosx \: }}$

$\boxed{\sf{ \:sin(x - y) = sinx \: cosy \: - \: siny \: cosx \: }}$

$\boxed{\sf{ \:cos(x + y) = cosx \: cosy \: - \: siny \: sinx \: }}$

$\boxed{\sf{ \:cos(x - y) = cosx \: cosy \: + \: siny \: sinx \: }}$

$\boxed{\sf{ \: \: tan(x + y) \: = \: \frac{tanx \: + \: tany}{1 \: - \: tanx \: tany \: } \: \: }}$

$\boxed{\sf{ \: \: tan(x - y) \: = \: \frac{tanx \: - \: tany}{1 \: + \: tanx \: tany \: } \: \: }}$

Answer 2

Step-by-step explanation:

$\tt \color{red}given : tan 9 \theta - tan 5 \theta - tan 4 \theta = tan 9 \theta \: tan5 \theta \: tan 4 \theta$

$\red{\boxed{ \overline{ \underline{ \tt \color{darkblue}solution : - }}}}$

$\tt \color{darkred} \: \: 9 \theta = 5 \theta + 4\theta$

$\tt \color{orange}take \: \: tan \: \: on \ \: both \: \: sides \:$

$\tt \color{navy} \tan(9\theta) = \tan(5\theta + 4\theta)$

$\tt \color{lime} \tan(9\theta) = \cfrac{ \tan(5\theta) \tan(4\theta) }{1 - \tan(5\theta) \tan(4\theta) }$

$\tt \color{green} \tan(9\theta) (1 - \tan(5\theta) \tan(4\theta) ) = \tan(5\theta) \tan(4\theta)$

$\tt \color{purple} \tan(9\theta) - \tan(5\theta) - \tan(4\theta) = \tan(9\theta) \tan(5\theta) \tan(4\theta)$

$\color{maroon}\mathbb{ HENCE \: \: \: PROVED..}$